Question

Find the sum of integers from $1$ to $100$ that are divisible by $2$ or $5$.

Answer 1

The integers between $1$ to $100$ which are divisible by $2$ are $2,4,\mathrm{6...},100$.
This forms an A.P. with both the first term and common difference equal to $2$.
$\Rightarrow 100=2+(n-1)2\Rightarrow n=50\therefore 2+4+6+...+100=\frac{50}{2}\left[2\right(2)+(50-1\left)\right(2\left)\right]=\frac{50}{2}[4+98]=\left(25\right)\left(102\right)=2550$
The integers from $1$ to $100$ which are divisible by $5$ are $5,10,..,100.$
This forms an A.P. with both the first term and common difference equal to $5$.
$\therefore 100=5+(n-1)5\Rightarrow 5n=100\Rightarrow n=20\therefore 5+10+...+100=\frac{20}{2}\left[2\right(5)+(20-1\left)5\right]=10[10+(19\left)5\right]=10[10+95]=10\times 105=1050$
The integers from $1$ to $100$ which are divisible by both $2$ and $5$ are $10,20,..,100$.
This forms an A.P. with both the first term and common difference equal to $10$.
$\therefore 100=10+(n-1)\left(10\right)\Rightarrow n=10\therefore 10+20+...+100=\frac{10}{2}\left[2\right(10)+(10-1\left)\right(10\left)\right]=5[20+90]=5\left(110\right)=550\therefore \text{Required sum}=2550+1050-550=3050$